Sum, sum, sum

let the terms be a, ar, ar^2 then a + ar + ar^2 = 14 .....(1) (ar + 1) - (a + 1) = (ar^2 - 1) - (ar + 1) ar + 1 - a - 1 = ar^2 - 1 - ar - 1 ar^2 - 2ar + a = 2 .......(2) (1) - (2) 3ar = 12 ar = 4 it follows that a = 2, ar = 4, ar^2 = 8 => r 2 or 1/2 the biggest term is 8.

The terms of the geometric progression are $a_1,a_2$ and $a_3$ . We let the terms of the arithmetic progression be $x_1,x_2$ and $x_3$ . Based form the problem,

$x_1 = a_1 + 1$

$x_2 = a_2 + 1$

$x_3 = a_3 - 1$

We know that the common difference in an arithmeric progression is equal. So

$x_2 - x_1 = x_3 - x_2$

Substituting, we have

$a_2 + 1 - (a_1 + 1) = a_3 - 1 - (a_2 + 1)$

$a_2 + 1 - a_1 - 1 = a_3 - 1 - a_2 - 1$

$a_2 - a_1 = a_3 - a_2 - 2$

$2a_2 + 2 = a_1 + a_3$ (equation 1)

From here, we are stuck in our solution, but in the problem it says that the sum of the geometric progression is $14$ , so we can use that information.

$a_1 + a_2 + a_3 = 14$

$a_1 + a_3 = 14 - a_2$ (equation 2)

We now equate the two equations.

$a_1 + a_3 = a_1 + a_3$

$2a_2 + 2 = 14 - a_2$

$3a_2 = 12$

$a_2 = 4$

Now our problem is $a_1$ and $a_3$ , we know the the common ratio in a geometric progession is equal. So

$\frac{a_2}{a_1} = \frac{a_3}{a_2}$

$\frac{4}{a_1} = \frac{a_3}{4}$

Cross multiply !

$16 = (a_1)(a_3)$ (equation 3)

From equation 2,

$a_1 + a_3 = 14 - a_2$

$a_1 + a_3 = 14 - 4$

$a_1 = 10 - a_3$ , substitute this to equation 3, we have

$16 = (a_1)(a_3)$

$16 = (10 - a_3)(a_3)$

$(a_3)^2 - 10a_3 + 16 = 0$

By factoring, we obtain

$a_3 = 8$ or $a_3 = 2$ , but $a_3$ must be larger that $a_2!$ so $a_3$ is $\boxed{8}$ .